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Question Number 90955 by jagoll last updated on 27/Apr/20

y′′−5y′−24y=e^(3x)

$${y}''−\mathrm{5}{y}'−\mathrm{24}{y}={e}^{\mathrm{3}{x}} \\ $$$$ \\ $$

Commented by jagoll last updated on 27/Apr/20

i got y_h = Ae^(8x) +Be^(−3x) y_p = ∫ e^(3x) dx = (1/3)e^(3x) general solution y=Ae^(8x) +Be^(−3x) +(1/3)e^(3x) it correct ?

$${i}\:{got}\: \\ $$$${y}_{{h}} \:=\:{Ae}^{\mathrm{8}{x}} +{Be}^{−\mathrm{3}{x}} \\ $$$${y}_{{p}} \:=\:\int\:{e}^{\mathrm{3}{x}} \:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{3}}{e}^{\mathrm{3}{x}} \\ $$$${general}\:{solution} \\ $$$${y}={Ae}^{\mathrm{8}{x}} +{Be}^{−\mathrm{3}{x}} +\frac{\mathrm{1}}{\mathrm{3}}{e}^{\mathrm{3}{x}} \\ $$$${it}\:{correct}\:? \\ $$

Commented by jagoll last updated on 27/Apr/20

what wrong my way? can anyone explain me?

$${what}\:{wrong}\:{my}\:{way}?\:{can}\:{anyone} \\ $$$${explain}\:{me}? \\ $$

Commented by peter frank last updated on 27/Apr/20

correct

$${correct} \\ $$

Commented by jagoll last updated on 27/Apr/20

are you sure? my friend say it not correct the answer in last term −(e^(3x) /(30))

$${are}\:{you}\:{sure}?\:{my}\:{friend}\:{say}\: \\ $$$${it}\:{not}\:{correct} \\ $$$${the}\:{answer}\:{in}\:{last}\:{term}\: \\ $$$$−\frac{{e}^{\mathrm{3}{x}} }{\mathrm{30}} \\ $$

Commented by john santu last updated on 27/Apr/20

•particular solution y_p (x)= pe^(3x) { ((y_p ′(x)= 3pe^(3x) )),((y_p ′′(x)= 9pe^(3x) )) :} comparing coeff lhs and rhs y_p ′′(x)−5y′_p (x)−24y_p = e^(3x) 9p−15p−24p = 1 p= −(1/(30)). ⇒y_p = −(1/(30))e^(3x) general solution ∴ y = Ae^(8x) + Be^(−3x) −(1/(30))e^(3x)

$$\bullet\mathrm{particular}\:\mathrm{solution} \\ $$$$\mathrm{y}_{\mathrm{p}} \left({x}\right)=\:{pe}^{\mathrm{3}{x}} \:\begin{cases}{{y}_{{p}} '\left({x}\right)=\:\mathrm{3}{pe}^{\mathrm{3}{x}} }\\{{y}_{{p}} ''\left({x}\right)=\:\mathrm{9}{pe}^{\mathrm{3}{x}} }\end{cases} \\ $$$$\mathrm{comparing}\:\mathrm{coeff}\:\mathrm{lhs}\:\mathrm{and}\:\mathrm{rhs} \\ $$$$\mathrm{y}_{\mathrm{p}} ''\left({x}\right)−\mathrm{5y}'_{\mathrm{p}} \left({x}\right)−\mathrm{24y}_{\mathrm{p}} =\:\mathrm{e}^{\mathrm{3}{x}} \\ $$$$\mathrm{9}{p}−\mathrm{15}{p}−\mathrm{24}{p}\:=\:\mathrm{1} \\ $$$${p}=\:−\frac{\mathrm{1}}{\mathrm{30}}.\:\Rightarrow\mathrm{y}_{\mathrm{p}} \:=\:−\frac{\mathrm{1}}{\mathrm{30}}\mathrm{e}^{\mathrm{3}{x}} \\ $$$$\mathrm{general}\:\mathrm{solution}\: \\ $$$$\therefore\:\mathrm{y}\:=\:\mathrm{Ae}^{\mathrm{8}{x}} \:+\:{B}\mathrm{e}^{−\mathrm{3}{x}} \:−\frac{\mathrm{1}}{\mathrm{30}}{e}^{\mathrm{3}{x}} \\ $$

Commented by jagoll last updated on 27/Apr/20

waw..thank you sir john

$${waw}..{thank}\:{you}\:{sir}\:{john} \\ $$

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